Calculating Baseball Parameters from Experimental Data

An engineering professor has a digital camera that takes short videos by taking a series of still pictures at the rate of 5 frames per second. He takes such a video of a ball being batted, and he is quite sure that the trajectory is perpendicular to his line of sight so that his camera records the arc as it actually is in two dimensions. He wants to see if he can determine the initial velocity of the ball and its initial angle by examining a series of still frames. By superimposing a grid on the frames, he determines the x and y coordinates of the

ball at 19 points in its trajectory. Because the professor had a map of the field he was able to scale the photos to the actual dimensions.

Using all 19 selected frames, he constructed the following table of ball coordinates:

x / 0 / 5 / 9 / 14 / 19 / 24 / 28 / 33 / 38 / 42 / 47 / 52 / 57 / 61 / 66 / 71 / 75 / 80 / 85
y / 1 / 3 / 6 / 7 / 9 / 10 / 11 / 12 / 12 / 13 / 12 / 12 / 11 / 11 / 9 / 8 / 6 / 5 / 2

He then plotted the data and used a spreadsheet to determine the best fit second-order polynomial, from which he calculated the initial angle and the initial velocity of the ball, as shown below. What is the maximum height? What is the range? How long does it take for the ball to land?

Using Excel he determined that the best fit second-order polynomial is:

yo = 1.0585 + 0.5232x – 0.006x2

tan o = 0.5232 o = 27.6o cos o = 0.886 (cos o)2 = 0.785

4.905/(vo2 * 0.785) =0.006 à vo2 = 1041

vo = 32 m/s vx = 32*cos 27.6o = 28.3 m/s

From the graph, the maximum height is about 12.5 meters. From 0 = 1.0585 + 0.5232x – 0.006x2 , x = Range = 89 meters, which compares well with the curve extended to the x axis. X distance traveled in 18 intervals * 0.2 s/interval = 3.6 s. x distance traveled in 3.6 s = 85 meters. Time to travel 89 meters is: 3.6*89/85 = 3.8 s. Or 89 m = 28.3*t à t = 3.14 s.